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CBSE Questions for Class 11 Commerce Applied Mathematics Numerical Applications Quiz 12 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Numerical Applications
Quiz 12
If four men working at the same rate can do $$2/3$$ of a job in $$40$$ minutes it take one man working at this rate to do $$2/5$$ of the job takes how many minutes?
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$$80$$
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$$88$$
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$$92$$
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$$96$$
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$$112$$
A library has $$'a'$$ copies of one book, $$'b'$$ copies of each of two books, $$'c'$$ copies of each of three books, and single copy each of $$'d'$$. The total number of ways in which these books can be arranged in a row is
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$$\dfrac{(a+b+c+d)!}{a!b!c!}$$
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$$\dfrac{(a+2b+3c+d)!}{a!(b!)^{2}(c!)^{3}}$$
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$$\dfrac{(a+2b+3c+d)!}{a!b!c!}$$
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none of these
$$\text{The number of ways dividing 52 cards amongst four players equally, are} $$
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$$\dfrac {52!}{(13!)^{4}}$$
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$$\dfrac {52!}{(13!)^{4}4!}$$
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$$\dfrac {52!}{(12!)^{4}4!}$$
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$$\text{None of these}$$
Explanation
$$\textbf{Hint:-Number of ways in which 3n diiferent things can be divided equally three people }$$
$$=\large\bf\frac{(3n)!}{n!n!n!}$$
$$\textbf{Step:-1 Find the total number of ways }$$
$$\text{Here 52 cards are distributed among four players equally in an order so each gets 13 cards}$$
$$\text{Total number of ways dividing 52 cards amongst four players equally,are=}$$ $$^{52}C_{13}\ \times \ ^{39}C_{13}\ \times$$$$\ ^{26}C_{13}\ \times \ ^{13}C_{13}$$
$$=\large\frac{52!}{39!13!}$$$$\times\large\frac{39!}{26!13!}$$$$\times\large\frac{26!}{13!13!}$$$$\times\large\frac{13!}{0!13!}$$
$$=\large\frac{52!}{13!\times13!\times13!\times13!}$$
$$=\large\frac{52!}{{13!}^4}$$
$$\textbf{Heence the number of ways the cards can be divided equally amongst number of players=}$$$$\large\bf\frac{52!}{{13!}^4}$$
If a pipe can fill a cistern in 8 hours and another pipe can empty the cistern in 10 hours, then what part of the cistern will be filled in 1 hour, if both pipes are opened simultancously?
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$$\frac { 1 }{ 40 } $$
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$$\frac { 1 }{ 30 } $$
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$$\frac { 1 }{ 35 } $$
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$$\frac { 1 }{ 8 } $$
A pipe can fill a cistern in $$12$$ min and another pipe can fill it in $$15$$ min, but a third pipe can empty it in $$6$$ min. The first two pipes are kept open for $$5$$ min in the beginning and then the third pipe is also opened. Time taken to empty the cistern is
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30 min
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33 min
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$$37\frac{1}{2}\min $$
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45 min
A street-light is hung $$20\ ft$$, above the ground. An object falls freely under the gravity, starting from rest at the same height as the lamp and at a horizontal distance of $$5\ ft$$. from it. When the object has fallen through $$16\ ft$$, the speed of the shadow of the object on the ground is
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$$12\ ft./sec$$
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$$11\ ft./sec$$
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$$10\ ft./sec$$
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$$12.5\ ft./sec$$
Four of the six numbers 1867, 1993, 2019, 2025, 2109 and 2121 have a mean ofWhat is the mean of the other two numbers.
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$$1994$$
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$$2006$$
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$$2022$$
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$$2051$$
Two pipes fill a tank in 20 minutes and 60 minutes. A third pipe empties the same tank in 40 minutes. In how much time will the tank be filled if all three pipes are opened together?
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6 min
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24 min
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30 min
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40 min
I
f a pipe fills a tank in 20 minutes and a pipe empties the same tank in 60 minutes. Then in how much time the tank will be filled completely if both the pipes are opened together?
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10 minutes
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70 minutes
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40 minutes
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30 minutes
The mean of series $$ x_{1},x_{2},x_{3},........x_{n} \ is \bar{x},$$ then mean of the series $$x_{1},+2i, i=1,2,3,....................n$$ will be
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$$ \bar{x}+n$$
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$$ \bar{x}+n+1 $$
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$$ \bar{x}+2$$
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$$ \bar{x}+2n$$
If $$16$$ men can build a wall of $$52$$m long in $$25$$ days working for $$8$$ hours a day, in how many days can $$64$$ men build a similar wall of $$260$$m long working for $$10$$ hours a day?
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$$35$$
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$$65$$
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$$40$$
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$$25$$
Explanation
$$\text { Solve:- on considering first situation } \\$$
$$\text { no. of men }\left(m_{1}\right)=16 \\$$
$$\text { no. of } \operatorname{days}\left(D_{1}\right)=25 \\$$
$$\text { no. of hours work daily }=8 \text { hours } \\$$
$$\text { - length of wall }=52 \mathrm{~m}$$
and in second case by using similar
notation we get,
$$M_{2} =64 \\ D_{2} =?$$
no. of hour $$=10$$ hours
length of wall $$=260 \mathrm{~m}$$
We know that, no. of men $$\alpha$$ work
no. of hours $$\alpha$$ work
no. of days $$\alpha$$ work
$$\text { So, } \frac{M_{1} \times D_{1} \times H_{1}}{l_{1}}=\frac{M_{2} \times D_{2} \times H_{2}}{l_{2}} \\$$
$$ \Rightarrow \frac{16 \times 25 \times 8}{52}=\frac{64 \times D_{2} \times 10}{260} $$
$$\quad \Rightarrow D_{2}=25$$ Days
$$H_{1}, H_{2}$$ Denotes hours in first and second
case. Similarly $$l_{1}$$ and $$l_{2}$$ represent length
A and $$B$$ can do a job together in $$7$$ days $$A$$ is $$1\dfrac { 3 } { 4 }$$ times as efficient as $$B$$ The same
job can be done by $$A$$ alone in
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$$9 \dfrac { 1 } { 3 }days$$
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$$11 days$$
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$$12\dfrac { 1 } { 4 } days$$
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$$16\dfrac { 1 } { 2 } days$$
Explanation
$$A$$ and $$B$$ can do a job together in 7 days $$A$$ is $$1 \frac{3}{4}$$ times as efficient as $$B$$
$$ \begin{aligned} \Rightarrow \quad A: B &=\mid \frac{3}{4}: 1 \\ &=7: 4 \\ A &=7 x \quad, \quad B=4 x \end{aligned} $$
$$\underline{A \cdot T \cdot Q}$$
$$ \begin{array}{l} A+B=\frac{1}{7} \\ 7 x+4 x=\frac{1}{7} \\ \Rightarrow x=\frac{1}{77} \\ \therefore A=7 x=\frac{1}{11}=11 \text { days } \end{array} $$
Hence, (B) is the correct option.
In a conference hall 10 executive are to be seated on 10 chairs placed left to right. Executive $$E_1$$ wants to sit to the right of $$E_2$$ Also, $$E_4$$ wants to sit the left of $$E_1$$ and $$E_5$$ wants to sits to the left of ways $$E_4$$ Number of ways the executives can be seated on the 10 chairs, is
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$$\dfrac{10!}{4!}$$
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$$\dfrac{10!}{3}$$
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$$10!$$
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$$3 \times \dfrac{10!}{4!}$$
Number of positive integers $$n ,$$ less than $$17 ,$$ for which $$n ! + ( n + 1 ) ! + ( n + 2 ) !$$ is an integral multiple of $$49$$ is
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$$2$$
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$$5$$
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$$7$$
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$$9$$
$$4$$ men and $$3$$ women finish a job in $$6$$ days. And $$5$$ men and $$7$$ women can do the same job in $$4$$ days. How long will $$1$$ man and $$1$$ woman take to do the work?
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$$22\dfrac { 2 }{ 7 } days$$
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$$25\dfrac { 1 }{ 2 } days$$
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$$5\dfrac { 1 }{ 7 } days$$
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$$12 \dfrac { 7 }{ 22 } days$$
Explanation
($$4$$ men $$+3$$ women)$$\times 6$$ days$$=$$($$5$$ men $$+ 7$$ women)$$\times 4$$ days
$$24$$ men $$+18$$ women$$=20$$ men $$+28$$ women
$$4$$ men days $$=10$$ women days
$$1$$ men days $$=2.5$$ women days
Total work $$=24$$ men days $$+18$$ women days
$$=20$$ men days $$+28$$ women days
$$=78$$ women days
$$1$$ man $$+1$$ women $$=2.5$$ women $$+1$$ women $$=3.5$$
women can do $$\dfrac{78}{3.5}=22\left(\dfrac{2}{7}\right)$$ days
If $$m$$ and $$n$$ are positive integers more than or equal to $$2 , m > n ,$$ then $$( m n ) !$$ is divisible by
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$$( m ! ) ^ { n }$$
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$$( n ! ) ^ { m }$$
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$$( m + n ) !$$
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$$( m - n ) !$$
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